I was watching a demonstration of the relation $$\left( \frac{x+1}{2} \right)^x>x!$$ for all $x > 1$
And I thought about the biggest possible denominator in the LHS such that it will be eventually bigger than the RHS, that is, for which values of $k$ does $$\left( \frac{x+1}{k} \right)^x>x!$$ hold for some $n<x$ sufficiently large.
By experimentally testing with WolframAlpha, it seems that if $k>e$ there is no such $n$.
Is there a way to prove it?
Better than the Stirling approximation, this interesting work by Milan Merkle provides a bilateral approximation to $x!$ which reads $$ \sqrt {2\,\pi } \;{{\left( {x + n} \right)^{\,\,x + n - 1/2} \;e^{\, - \;\left( {x + n} \right)} } \over {x^{\,\overline {\,n\,} } }}\;\; \le \;\;\Gamma (x)\; \le \;\sqrt {2\,\pi } \;{{\left( {x + n} \right)^{\,x + n - 1} \sqrt {\,x + n + 1/2} \;e^{\, - \;\left( {x + n} \right)} } \over {x^{\,\overline {\,n\,} } }}\quad \quad \left| \matrix{ \;0 < x \hfill \cr \;0 \le n\; \to \infty \hfill \cr} \right. $$ where $x^{\,\overline {\,k\,} } $ represent the Rising Factorial.
We can rewrite the above as $$ \sqrt {{{2\,\pi } \over e}} \;\left( {{{x + n} \over e}} \right)^{\,\,x + n - 1/2} {{\;1} \over {x^{\,\overline {\,n\,} } }}\;\; \le \;\;\Gamma (x)\; \le \;\sqrt {{{2\,\pi } \over e}} \;\left( {{{x + n} \over e}} \right)^{\,\,x + n - 1/2} {{\;1} \over {x^{\,\overline {\,n\,} } }} \sqrt {\,1 + {{1/2} \over {x + n}}} \quad \left| \matrix{ \;0 < x \hfill \cr \;0 \le n\; \to \infty \hfill \cr} \right. $$ i.e. $$ \sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1 + n} \over e}} \right)^{\,\,x + n + 1/2} {{\;1} \over {\left( {x + 1} \right)^{\,\overline {\,n\,} } }}\;\; \le \;\;x!\; \le \;\sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1 + n} \over e}} \right)^{\,\,x + n + 1/2} {{\;1} \over {\left( {x + 1} \right)^{\,\overline {\,n\,} } }}\sqrt {\,1 + {{1/2} \over {x + 1 + n}}} \quad \left| \matrix{ \;0 < x \hfill \cr \;0 \le n\; \to \infty \hfill \cr} \right. $$ and taking $n=0$ $$ \eqalign{ & \sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1} \over e}} \right)^{\,\,x + 1/2} \;\; \le \;\;x!\; \le \;\sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1} \over e}} \right)^{\,\,x + 1/2} \sqrt {\,1 + {{1/2} \over {x + 1}}} \quad \left| {\;0 < x} \right. \cr & 1\;\; \le \;\;{{x!} \over {\sqrt {{{2\,\pi } \over e}} \;\left( {{{x + 1} \over e}} \right)^{\,\,x + 1/2} }}\; \le \;\sqrt {\,1 + {{1/2} \over {x + 1}}} \quad \left| {\;0 < x} \right. \cr} $$